Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in PYTHON):

Solution:
#!/usr/bin/env python2
"""
This file is part of https://github.com/cheran-senthil/PyRival
Copyright 2019 Cheran Senthilkumar <hello@cheran.io>

"""
from __future__ import division, print_function

import itertools
import math
import os
import sys
from atexit import register
from io import BytesIO


class dict(dict):
    """dict() -> new empty dictionary"""

    def items(self):
        """D.items() -> a set-like object providing a view on D's items"""
        return dict.iteritems(self)

    def keys(self):
        """D.keys() -> a set-like object providing a view on D's keys"""
        return dict.iterkeys(self)

    def values(self):
        """D.values() -> an object providing a view on D's values"""
        return dict.itervalues(self)


def gcd(x, y):
    """greatest common divisor of x and y"""
    while y:
        x, y = y, x % y
    return x


range = xrange

filter = itertools.ifilter
map = itertools.imap
zip = itertools.izip

sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
sys.stdout = BytesIO()
register(lambda: os.write(1, sys.stdout.getvalue()))

input = lambda: sys.stdin.readline().rstrip('\r\n')

n = int(input())
p = [float(i) for i in input().split()]

dp = [[0.0] * (n + 1) for _ in range(n)]
dp[0][0], dp[0][1] = 1 - p[0], p[0]

for i in range(1, n):
    dp[i][0] = (1 - p[i]) * dp[i - 1][0]
    for j in range(1, n + 1):
        dp[i][j] = (p[i] * dp[i - 1][j - 1]) + ((1 - p[i]) * dp[i - 1][j])

print(sum(dp[-1][(n + 1) // 2:]))